# Solve $e^x+x=1$

This seems to have stumped even my TA, so I'm asking it here.

Given $e^x + x = 1$, solve for $x$.

I already know that the answer is zero, but have no idea how to get there.

• Under most reasonable meanings of algebraic, no. Exponentials $+$ polynomials $=$ mess. – André Nicolas Sep 28 '12 at 18:40
• You might want take a look at Lambert's W. – EuYu Sep 28 '12 at 18:52
• I'm teaching from Stewart this term and assigned this problem as homework. The point is that if you have an invertible function $f$ such that $f(0) = 1$ (such as the function $f(x) = e^x + x$), then you know that $f^{-1}(1) = 0$, even if you can't solve the equation $f(x) = a$ for $x$ algebraically in terms of $a$. – Michael Joyce Sep 28 '12 at 19:56
• @MichaelJoyce And, we know it's invertible because it is increasing an increasing function. – GeoffDS Sep 28 '12 at 21:52
• @StevenStadnicki $f(x)e^x$ is well known to be increasing is not an "algebraic" argument ;) My point is that technically the exponential function is hard to define, let alone study, purely algebraically, you need to use some analysis some way or another.... – N. S. Aug 18 '14 at 18:47

Using the series expansion we have:

$$1+x+\frac{x^2}{2!}+ \dots + \frac{x^n}{n!}+\dots = 1-x$$

If $x$ is positive it is immediately obvious that there can be no equality.

If $x<0$ then the RHS is greater than 1 and $e^{x}<1$.

This is not strictly an "algebraic" solution, but with the term in $e^x$ we do not expect anything purely algebraic.

• 0.o Good enough for my purposes. Thanks! – YellPika Sep 28 '12 at 18:58

"Lambert W" is a hint for "algebraic solution".
The solution for $\mathrm{e}^x + x = 1$ is $1-\mathrm W(\mathrm{e})$,
to find ALL complex solutions, use all branches of the Lambert W ...

\begin{align*} &\dots \\ 1 - \mathrm{W}_{-4}(\mathrm{e}) &= 3.159947300 + 23.47017395 i \\ 1 - \mathrm{W}_{-3}(\mathrm{e}) &= 2.849014724 + 17.17149358 i \\ 1 - \mathrm{W}_{-2}(\mathrm{e}) &= 2.393982241 + 10.86800606 i \\ 1 - \mathrm{W}_{-1}(\mathrm{e}) &= 1.532092122 + 4.597158013 i \\ 1 - \mathrm{W}_{0}(\mathrm{e}) &= 0.000000000 \\ 1 - \mathrm{W}_{1}(\mathrm{e}) &= 1.532092122 - 4.597158013 i \\ 1 - \mathrm{W}_{2}(\mathrm{e}) &= 2.393982241 - 10.86800606 i \\ 1 - \mathrm{W}_{3}(\mathrm{e}) &= 2.849014724 - 17.17149358 i \\ 1 - \mathrm{W}_{4}(\mathrm{e}) &= 3.159947300 - 23.47017395 i \\ 1 - \mathrm{W}_{5}(\mathrm{e}) &= 3.396557044 - 29.76478701 i \\ &\dots \end{align*}

explanation

$\mathrm{e}^x+x=1$
$\mathrm{e}^x=1-x$
$\mathrm{e} = (1-x)\mathrm{e}^{1-x}$
$\mathrm{W}(\mathrm{e}) = 1-x$
$x = 1-\mathrm W(\mathrm{e})$

You can see this very easily graphically. The equation is $$e^x=1-x$$ and the two sides of the equation are plotted here (from Wolfram Alpha):

The intuition for a formal proof also follows directly from the picture (the functions are both monotonic but in opposite directions), if that's your aim.

Let $f(x) = e^x + x - 1$. Then, for any given $x$, $f(x) = 0$ if and only if $e^x + x = 1$.

You have already noticed that $f(0) = 1 + 0 - 1 = 0$, so it is a solution.

Now, we turn to calculus, not algebra. We have $f'(x) = e^x + 1$. Since $e^x > 0$ for all $x$, we know that $e^x + 1 > 0$ as well. In other words, $f'(x)$ is positive for all $x$ which tells us that $f(x)$ is an increasing function on the entire real line. Therefore, it could only possibly be 0 at one point, and you already found that point.

Now, if you haven't had calculus, you could still get the same basic idea. For example, you know $y = x$ is increasing. That is something you should know. Perhaps you have learned that $y = e^x$ is always increasing as well, because even in an algebra class, they would probably give you a bunch of properties of $y = e^x$ when they introduce it. Add these two functions together, and it's still increasing. Subtract 1, and the function is simply translated downward 1 unit, so it's still increasing everywhere. Again, the conclusion is the same.

• Note that this is in its essence, Rolle's theorem – Dimitris Aug 30 '20 at 5:23

Just to add weight to @Jonathan’s response: If $f$ and $g$ are an increasing and a decreasing function on $\mathbb{R}$ respectively, their graphs can cross at only one point. Inspection finds that point to be $(0,1)$, and you’re done.

This problem can be solved using domain arguments:

$e^{x} + x = 1$
$e^{x} = 1 - x$
$ln(e^{x}) = ln(1 - x)$
$x\cdot ln(e) = ln(1 - x)$

$x = ln(1-x)$

• $x$ cannot be larger than one, because then the expression $1-x$ will be negative violating the domain of a logarithmic function.
• $x$ cannot be a positive number between 0 and 1, because $ln(1-x)$ will be a negative value
• For the same reason, $x$ cannot be a number less than zero because $ln(1-x)$ will be a positive value

The only possible value for $x$ is 0.

Trial By Error is the only real solution.

Working with the graphical solution above, we can see that we can start with any random value for x, then recursively calling:-

for(x=i=0; i<5; x=1-((1-x+pow(M_E,x))/2.0), i++);

This takes the average value for both functions at the value of x, then re-inserting the new value for f(x) back into one of the original functions, to get a new value for x. Repeated 5 or so times to get a close estimate for the true value of x.

• This question is an old question which has a well accepted answer. You have contributed nothing new – Shailesh Jan 9 '16 at 2:04
• Hello Shailesh, I guessing you do not understand the question or answer. Which previous answer is exploiting the average value between the function as an estimate for the result? – Dallas Clarke Jan 9 '16 at 2:09
• Since $x = 0$ is the only real solution, you are trying to arrive at it through numerical methods. That does not constitute as a good answer to the question. – Shailesh Jan 9 '16 at 2:32
• The solution can be used for all situations, not just the one in the question above. There is no mathematical solution to this problem, but this exploits some mathematical properties to produce an efficient enough solution. In my situation, looping 10 times produced more reliable results. – Dallas Clarke Mar 13 '16 at 0:45

f(x)=e^x+x-1 The obvious solution is 0 and you can prove its the only one. By differentiating you get f'(x)=e^x+1 which is always positive.That means the f(x) function can only have one solution for zero because its monotonic.

A semi-intuitive way to think about it: the right-hand side has no $e$, so whatever you plug into the $e^x$ has to kill the $e$. So it has to be $\ln a$ for some $a$. On the other hand, the right hand side has no $\ln$, so what you plug into the $x$ term must reduce to a non-$\ln$. If you try $\ln e^b$, you'll end up in a circle. If you think maybe things will cancel, you'll see that it wont work because you will always have a different number of "levels" of exponentials and logarithms. The only possibility is $\ln 1 = 0$, which quick inspection reveals that it works.

This is by no means a formal proof (at least as I've written it), but it provides the intuition to find the solution $x=0$ if you failed to notice it off the bat.

This also shows why the solution to something like $e^x + x = 2$ is not going to expressible in terms of functions like exponentials and natural logs.

And by the way, if you are also interested in complex solutions, all bets are off with this method, since $e^{x+2\pi i}=e^x$.

Let's keep this one simple...

$$e^x+x=1$$

What here can give you $1+0$? Well, we know anything raised to 0 equals 1, and anything plus 0 equals it's self (at least in the reals) so... $e^0 + 0 = 1 + 0 = 1$... Just simple math intuition makes it easy to see that $e^x+x=1$ when $x=0$.

• Hi, welcome to Math stack exchange. To begin, maybe help us to understand your post by editing the equation? Please use LaTeX as this is the language we use here in the community. – bryanblackbee Apr 12 '15 at 17:21
• @Barry, I have improved the math formatting in your answer but you should try to improve the content of your answer. – Prasun Biswas Apr 20 '15 at 1:47